Question: You have found the following ages (in years) of all 5 zebras at your local zoo: $ 9,\enspace 9,\enspace 31,\enspace 29,\enspace 15$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{9 + 9 + 31 + 29 + 15}{{5}} = {18.6\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $9$ years $-9.6$ years $92.16$ years $^2$ $9$ years $-9.6$ years $92.16$ years $^2$ $31$ years $12.4$ years $153.76$ years $^2$ $29$ years $10.4$ years $108.16$ years $^2$ $15$ years $-3.6$ years $12.96$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{92.16} + {92.16} + {153.76} + {108.16} + {12.96}} {{5}} $ $ {\sigma^2} = \dfrac{{459.2}}{{5}} = {91.84\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{91.84\text{ years}^2}} = {9.6\text{ years}} $ The average zebra at the zoo is 18.6 years old. There is a standard deviation of 9.6 years.